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haleyelizabeth2017

  • one year ago

The area of a parking lot is 805 square meters. A car requires 5 square meters and a bus requires 32 square meters of space. There can be at most 80 vehicles parked at one time. If the cost to park a car is $2.00 and a bus is $6.00, how many buses should be in the lot to maximize income? Please no direct answers.

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  1. haleyelizabeth2017
    • one year ago
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    Would the first inequality be \[5c+32b \le 805\]?

  2. Hero
    • one year ago
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    looks right

  3. haleyelizabeth2017
    • one year ago
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    Okay...I'm not sure what to put for the second one...\(c+b\le 80\)?

  4. Hero
    • one year ago
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    why did you write 805 Instead of 80 ?

  5. haleyelizabeth2017
    • one year ago
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    there are 805 square meters in the parking lot.

  6. haleyelizabeth2017
    • one year ago
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    Shoot...typo, sorry.

  7. Hero
    • one year ago
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    Wait. Yoo mean 85

  8. haleyelizabeth2017
    • one year ago
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    I made a typo...I meant 805, my bad

  9. Hero
    • one year ago
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    Assuming your inequalities are correct, what do you think would be the next step?

  10. haleyelizabeth2017
    • one year ago
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    Don't I need a third inequality? For prices?

  11. haleyelizabeth2017
    • one year ago
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    oh wait...no...I use the prices to find the min/max! So f(x,y)=2c+6b?

  12. Hero
    • one year ago
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    Correct

  13. haleyelizabeth2017
    • one year ago
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    Will there only be one intersection point then?

  14. haleyelizabeth2017
    • one year ago
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    But, would max be at (65,15) so 15 buses?

  15. Hero
    • one year ago
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    There's a range of points Within a region that are possible, but there is only one correct pair of coordinates that will maximize the profit

  16. Hero
    • one year ago
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    Show we how you found that result.

  17. haleyelizabeth2017
    • one year ago
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    I have no idea...I just chose the intersection point...I'm sorry, I'm having a hard time understanding these word problems...with only two inequalities to graph...

  18. Hero
    • one year ago
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    There's almost always four inequalities to graph and you usually use x and y for variables

  19. haleyelizabeth2017
    • one year ago
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    When I dealt with the three inequalities I plugged in the intersection points into the f(x,y) function...

  20. Hero
    • one year ago
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    Because you also consider .\[c \ge 0\] and \[b \ge 0\]

  21. haleyelizabeth2017
    • one year ago
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    oh.

  22. Hero
    • one year ago
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    but If you use x and y as variables, then you can use a platform like desmos to plot your inequalities

  23. haleyelizabeth2017
    • one year ago
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    That's what I'm doing :)

  24. haleyelizabeth2017
    • one year ago
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    So, the intersection points I have now are: (80, 0) (65, 15) (0, 25.16) (0,0) But we can tell already that (0,0) will not be the maximum.

  25. haleyelizabeth2017
    • one year ago
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    It is (65,15).

  26. haleyelizabeth2017
    • one year ago
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    f(80,0)=160 f(0, 25.16)=150.96 f(0,0)=0 f(65,15)=220

  27. haleyelizabeth2017
    • one year ago
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    Would you mind helping me with a couple more?

  28. Hero
    • one year ago
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    Sure, but you seem to already know what you are doing for the Most part.

  29. haleyelizabeth2017
    • one year ago
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    I think for one I just need it checked...not sure about the last one..

  30. haleyelizabeth2017
    • one year ago
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    Eh I'll just post the one I need help on :)

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